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NEW QUESTION: 1
다음 중 비즈니스 프로세스, 애플리케이션 및 데이터 저장소에 대한 데이터 엔드 링크를 식별하고 소유 책임을 지정하는 것은 무엇입니까?
A. 위험 평가
B. 보안 거버넌스
C. 보안 포트폴리오 관리
D. 위험 관리
Answer: D

NEW QUESTION: 2
James would like to import data from a text file that contains foreign symbols.
What kind of file will he need to use?
A. Language translation file (CSS)
B. Character conversion language file (COL)
C. File translation Definition file (CDS)
D. Character translation file (CLS)
Answer: D

NEW QUESTION: 3
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.0.255.255.255.254
B. 許可10.0.0.0.0.0.0.1
C. 許可10.0.0.1.0.0.0.0
D. 許可10.0.0.1.0.0.0.254
Answer: D
Explanation:
Explanation
Remember, for the wildcard mask, 1s are I DON'T CARE, and 0s are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum: 0001111 = 31).

The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered hosts in the 10.0.0.0/24 subnet.

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